Another ladder problem.

bobbym · 2012-10-26 23:41:40

Hi;

This came up in another thread:

A 12-foot ladder is leaning across a fence and is touching a higher wall located 3 feet behind the fence. The ladder makes an angle of 60 degrees with the ground. Find the distance from the base of the ladder to the bottom of the fence.

The OP solved it and it is not too difficult to do in a number of ways. If it is fairly easy using trig then it is smashingly simple using geogebra!

1) Start by drawing a slider called a on the top of the screen.

2) Set Min to 0 and Max to 8 and increment to .01.

3) Move the slider to the extreme right until is says a = 8.

4) Enter (a,0) to create point A. It will be located at (8,0).

5) Create point (0,0) to mark the bottom of the wall.

6) Use the angle with a given size tool and click B then A and enter 60° and clockwise. Angle alpha will be created and point B'.

7) Hide the angle and then draw a line through A and B'.

8) Hide B' and using the intersection tool find the point of intersection with that line and the y axis. Point C will be created.

9) Hide the line and draw line segment AC. This line segment represents the ladder.

10) The length of AC will be visible in the algebra pane. It is called c.

11) Move the slider until little c ( the length of the ladder ) equals 12.

12) Now the fence is 3 ft. in front of the wall. So create point D by entering (3,0).

13) Now draw a perpendicular line through D with the x axis. Get the point of intersection with that line and the ladder (AC). Point E will be created.

14) Hide the vertical line and draw line segment DE. This is the fence.

The base of the ladder is at point A (6,0) the base of the fence is at D (3,0). The distance is obviously 3 ft. We are done. Your drawing should look like the one below.

Math Is Fun Forum

#1 2012-10-26 23:41:40

Another ladder problem.

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